Introduction to Complex Analysiscomplex number z 0. There is an important diﬀerence between these...

Introduction to Complex Analysis

George Voutsadakis1

1Mathematics and Computer ScienceLake Superior State University

LSSU Math 413

George Voutsadakis (LSSU) Complex Analysis October 2014 1 / 45

Outline

1 Limits and ContinuityLimitsContinuity


Limits and Continuity Limits

Subsection 1

Limits



Real and Complex Limits

limx→x0 f (x) = L intuitively means that values f (x) of the function f

can be made arbitrarily close to the real number L if values of x arechosen sufficiently close to, but not equal to, the real number x0.

In real analysis, the concepts of continuity, the derivative, and thedefinite integral were all defined using the concept of a limit.

limz→z0 f (z) = L will mean that the values f (z) of the complexfunction f can be made arbitrarily close to the complex number L ifvalues of z are chosen sufficiently close to, but not equal to, thecomplex number z0.There is an important difference between these two concepts of limit:

In a real limit, there are two directions from which x can approach x0on the real line, from the left or from the right.In a complex limit, there are infinitely many directions from which z

can approach z0 in the complex plane. In order for a complex limit toexist, each way in which z can approach z0 must yield the samelimiting value.



Real Limits: From Intuition to Formalism

To rigorously define a real limit, we must formalize what is meant bythe phrases “arbitrarily close to” and “sufficiently close to”.

A precise statement should involve the use of absolute values since|a − b| measures the distance between a, b on the real number line.

The points x and x0 are close if |x − x0| is a small positive number.

Also, f (x) and L are close if |f (x)− L| is a small positive number.

We let the Greek letters ε and δ represent small positive real numbers.

The expression “f (x) can be made arbitrarily close to L” can be madeprecise by stating that for any real number ε > 0, x can be chosen sothat |f (x)− L| < ε.

We require that |f (x)− L| < ε whenever values of x are “sufficientlyclose to, but not equal to, x0”.

This means that there is some distance δ > 0 with the property that,if x is within distance δ of x0 and x 6= x0, then |f (x)− L| < ε.



Formal Definition of a Real Limit

Definition (Limit of a Real Function f (x))

The limit of f as x tends to x0 exists and is equal to L if, for every ε > 0,there exists a δ > 0, such that |f (x)− L| < ε when 0 < |x − x0| < δ.

The geometric interpretation is shown:

The graph of the function y = f (x) over theinterval (x0− δ, x0+ δ), excluding the point x0,lies between the lines y = L− ε and y = L+ ε.In the terminology of mappings, the interval(x0 − δ, x0 + δ), excluding the point x = x0, ismapped onto a set in the interval (L−ε, L+ε)on the y -axis.



Complex Limits

A complex limit is based on a notion of “close” in the complex plane.

Because the distance in the complex plane between two points z1 andz2 is given by the modulus of the difference of z1 and z2, the precisedefinition of a complex limit will involve |z2 − z1|.E.g., the phrase “f (z) can be made arbitrarily close to the complexnumber L” can be stated precisely: “for every ε > 0, z can be chosenso that |f (z)− L| < ε.

Since the modulus of a complex number is a real number, both ε andδ still represent small positive real numbers:

Definition (Limit of a Complex Function)

Suppose that a complex function f is defined in a deleted neighborhood ofz0 and suppose that L is a complex number. The limit of f as z tends to

z0 exists and is equal to L, written as limz→z0 f (z) = L, if, for every ε > 0,there exists a δ > 0, such that |f (z)− L| < ε whenever 0 < |z − z0| < δ.



Geometric Representation

The set of points w satisfying |w − L| < ε is called a neighborhood ofL, and consists of all points in the complex plane lying within, but noton, a circle of radius ε centered at the point L.

The set of points satisfying 0 < |z − z0| < δ is called a deletedneighborhood of z0 and consists of all points in the neighborhood|z − z0| < δ excluding the point z0.

If limz→z0 f (z) = L and if ε is any positive number, then there is adeleted neighborhood of z0 of radius δ, such that, for every z in thisdeleted neighborhood, f (z) is in the ε neighborhood of L:



Real One-Sided Limits

There is at least one very important difference between real andcomplex limits.

For real functions, limx→x0 f (x) = L if and only if limx→x+0f (x) = L

and limx→x−

0f (x) = L. Since there are two directions from which x can

approach x0 on the real line, the real limit exists if and only if thesetwo one-sided limits have the same value.

Example: Consider the real function f (x) =

{

x2, if x < 0x − 1, if x ≥ 0

.

The limit of f as x approaches 0 does not exist:

limx→0− f (x) = limx→0− x2 = 0, butlimx→0+ f (x) = limx→0+ (x − 1) = −1.



Criterion for the Nonexistence of a Limit

For limits of complex functions, z is allowed to approach z0 from anydirection in the complex plane, i.e., along any curve or path throughz0.

For limz→z0 f (z) to exist and to equal L, we require that f (z)approach the same complex number L along every possible curvethrough z0.

Criterion for the Nonexistence of a Limit

If f approaches two complex numbers L1 6= L2 for two different curves orpaths through z0, then limz→z0 f (z) does not exist.



Example: Nonexistence of a Limit

Example: Show that limz→0zzdoes not exist.

We show that this limit does not exist by finding two different ways ofletting z approach 0 that yield different values for limz→0

zz.

First, we let z approach 0 along the real axis. That is, we considercomplex numbers of the form z = x + 0i , where the real number x isapproaching 0. For these points we have:

limz→0

z

z= lim

x→0

x + 0i

x − 0i= lim

x→01 = 1.

On the other hand, if we let z approach 0 along the imaginary axis,then z = 0 + iy , where the real number y is approaching 0. For thisapproach we have:

limz→0

z

z= lim

y→0

0 + iy

0− iy= lim

y→0(−1) = − 1.

Since the two values are not the same, we conclude that limz→0zz

does not exist.George Voutsadakis (LSSU) Complex Analysis October 2014 11 / 45


Epsilon-Delta Proofs

Computing values of limz→z0 f (z) as z approaches z0 from differentdirections can prove that a limit does not exist, but cannot be used toprove that a limit does exist.

To prove that a limit exists we must use the definition directly.

This requires demonstrating that for every positive real number εthere is an appropriate choice of δ that meets the relevantrequirements.

Such proofs are commonly called “epsilon-delta proofs”.

Even for relatively simple functions, epsilon-delta proofs can be quitecomplicated.

We only show some easy examples of such proofs.



Example: An Epsilon-Delta Proof

Prove that limz→1+i (2 + i)z = 1 + 3i .

According to the definition, limz→1+i (2 + i)z = 1 + 3i , if, for everyε > 0, there is a δ > 0, such that |(2 + i)z − (1 + 3i)| < ε whenever0 < |z − (1 + i)| < δ. Proving that the limit exists requires that wefind an appropriate value of δ for a given value of ε. One way offinding δ is to “work backwards”. The idea is to start with theinequality: |(2 + i)z − (1 + 3i)| < ε and then use properties ofcomplex numbers and the modulus to manipulate this inequality untilit involves the expression |z − (1 + i)|.We first factor (2 + i) out of the left-hand side:

|2 + i | ·∣

∣

∣z − 1+3i

2+i

∣

∣

∣< ε. Because |2 + i | =

√5 and 1+3i

2+i= 1 + i , we

get:√5 · |z − (1 + i)| < ε or |z − (1 + i)| < ε√

5. This indicates that

we should take δ = ε√5.



Example: An Epsilon-Delta Proof (Cont’d)

We now present the formal proof: Given ε > 0, let δ = ǫ√5. If

0 < |z − (1 + i)| < δ, then we have |z − (1 + i)| < ε√5. Multiplying

both sides by |2 + i | =√5 we obtain: |2 + i | · |z − (1 + i)| <

√5 · ε√

5

or |(2 + i)z − (1 + 3i)| < ε. Therefore, |(2 + i)z − (1 + 3i)| < εwhenever 0 < |z − (1 + i)| < δ. So, by definition,limz→1+i (2 + i)z = 1 + 3i .



Real Multivariable Limits

We present a practical method for computing complex limits whichalso establishes an important connection between the complex limit off (z) = u(x , y) + iv(x , y) and the real limits of the real-valuedfunctions of two real variables u(x , y) and v(x , y).

Since every complex function is completely determined by the realfunctions u and v , the limit of a complex function can be expressed interms of the real limits of u and v .

Definition (Limit of the Real Function F (x , y))

The limit of F as (x , y) tends to (x0, y0) exists and is equal to the realnumber L if, for every ε > 0, there exists a δ > 0, such that|F (x , y)− L| < ε whenever 0 <

√

(x − x0)2 + (y − y0)2 < δ.

The expression√

(x − x0)2 + (y − y0)2 represents the distancebetween the points (x , y) and (x0, y0) in the Cartesian plane.



Properties of Limits

Using the definitions, we can prove that:

lim(x,y)→(x0 ,y0) 1 = 1,lim(x,y)→(x0 ,y0) x = x0,lim(x,y)→(x0 ,y0) y = y0.If lim(x,y)→(x0 ,y0) F (x , y) = L and lim(x,y)→(x0 ,y0) G(x , y) = M , then:

lim(x,y)→(x0,y0) cF (x , y) = cL, c a real constant,lim(x,y)→(x0,y0) (F (x , y)± G(x , y)) = L±M,lim(x,y)→(x0,y0) F (x , y) · G(x , y) = L ·M,

lim(x,y)→(x0,y0)F (x , y)

G(x , y)=

L

M, M 6= 0.



Limits Involving Polynomial Expressions

Example: Limits involving polynomial expressions in x and y can beeasily computed using these rules:

lim(x ,y)→(1,2)

(3xy2 − y)

= 3(lim(x ,y)→(1,2) x)(lim(x ,y)→(1,2) y)(lim(x ,y)→(1,2) y)

− lim(x ,y)→(1,2) y

= 3 · 1 · 2 · 2− 2= 10.

In general, if p(x , y) is a two-variable polynomial function, then

lim(x ,y)→(x0,y0)

p(x , y) = p(x0, y0).

If p(x , y) and q(x , y) are two-variable polynomial functions andq(x0, y0) 6= 0, then

lim(x ,y)→(x0,y0)

p(x , y)

q(x , y)=

p(x0, y0)

q(x0, y0).



Real and Imaginary Parts of a Limit

Theorem (Real and Imaginary Parts of a Limit)

Suppose that f (z) = u(x , y) + iv(x , y), z0 = x0 + iy0 and L = u0 + iv0.Then limz→z0 f (z) = L if and only if

lim(x ,y)→(x0,y0)

u(x , y) = u0 and lim(x ,y)→(x0,y0)

v(x , y) = v0.

The theorem reduces the computation of complex limits to thecomputation of a pair of real limits.

Example: Compute limz→1+i (z2 + i).

Since f (z) = z2 + i = x2 − y2 + (2xy + 1)i , we setu(x , y) = x2 − y2, v(x , y) = 2xy + 1 and z0 = 1+ i , i.e., x0 = 1 andy0 = 1. We next compute the two real limits:

u0 = lim(x ,y)→(1,1) (x2 − y2) = 12 − 12 = 0,

v0 = lim(x ,y)→(1,1) (2xy + 1) = 2 · 1 · 1 + 1 = 3.

Therefore, L = u0 + iv0 = 0 + i(3) = 3i , i.e., limz→1+i (z2 + i) = 3i .



Properties of Complex Limits

Theorem (Properties of Complex Limits)

Suppose that f and g are complex functions. If limz→z0 f (z) = L andlimz→z0 g(z) = M, then:

(i) limz→z0 cf (z) = cL, c a complex constant;

(ii) limz→z0 (f (z)± g(z)) = L±M;

(iii) limz→z0 f (z) · g(z) = L ·M, and

(iv) limz→z0

f (z)

g(z)=

L

M, provided M 6= 0.

We only prove part (i): Let f (z) = u(x , y) + iv(x , y), z0 = x0 + iy0,L = u0 + iv0, and c = a + ib. Since limz→z0 f (z) = L,lim(x ,y)→(x0,y0) u(x , y) = u0 and lim(x ,y)→(x0,y0) v(x , y) = v0. Thenlim(x ,y)→(x0,y0) (au(x , y)− bv(x , y)) = au0 − bv0 andlim(x ,y)→(x0,y0) (bu(x , y) + av(x , y)) = bu0 + av0.



Properties of Complex Limits (Cont’d)

We set f (z) = u(x , y) + iv(x , y), z0 = x0 + iy0, L = u0 + iv0, andc = a + ib. We then computedlim(x ,y)→(x0,y0) (au(x , y)− bv(x , y)) = au0 − bv0 andlim(x ,y)→(x0,y0) (bu(x , y) + av(x , y)) = bu0 + av0.However, note that

cf (z) = (a + ib)(u + iv)= (au − bv) + i(bu + av).

Thus, Re(cf (z)) = au(x , y)− bv(x , y) andIm(cf (z)) = bu(x , y) + av(x , y). Therefore,limz→z0 cf (z) = au0 − bv0 + i(bu0 + av0) = (a + ib)(u0 + iv0) = cL.

Many limits can now be computed starting from:

limz→z0 c = c , c a complex constant;limz→z0 z = z0.



Computing Limits I

Compute the limit limz→i(3 + i)z4 − z2 + 2z

z + 1.

limz→i z2 = limz→i z · z = (limz→i z)(limz→i z) = i · i = −1.

Similarly, limz→i z4 = i4 = 1. Using these limits, and the properties,

we obtain: limz→i ((3 + i)z4 − z2 + 2z) = (3 + i) limz→i z4 −

limz→i z2 + 2 limz→i z = (3 + i)(1)− (−1) + 2(i) = 4 + 3i , and

limz→i (z + 1) = 1 + i . Therefore, finally,

limz→i(3 + i)z4 − z2 + 2z

z + 1=

limz→i ((3 + i)z4 − z2 + 2z)

limz→i (z + 1)=

4 + 3i

1 + i.

After carrying out the division, we obtain

limz→i(3 + i)z4 − z2 + 2z

z + 1= 7

2 − 12 i .



Computing Limits II

Compute the limit limz→1+

√3i

z2 − 2z + 4

z − 1−√3i.

limz→1+√3i (z

2 − 2z + 4) = (1 +√3i)2 − 2(1 +

√3i) + 4 =

− 2 + 2√3i − 2− 2

√3i + 4 = 0, and

limz→1+√3i (z − 1−

√3i) = 1 +

√3i − 1−

√3i = 0. It appears that

we cannot apply the quotient rule since the limit of the denominatoris 0. However, in the previous calculation we found that 1 +

√3i is a

root of the quadratic polynomial z2 − 2z + 4. If z1 is a root of aquadratic polynomial, then z − z1 is a factor of the polynomial. Usinglong division, we find that z2 − 2z + 4 = (z − 1 +

√3i)(z − 1−

√3i).

Because z is not allowed to take on the value 1 +√3i in the limit:

limz→1+√3i

z2 − 2z + 4

z − 1−√3i

= limz→1+√3i

(z − 1 +√3i)(z − 1−

√3i)

z − 1−√3i

=

limz→1+√3i (z − 1 +

√3i) = 1 +

√3i − 1 +

√3i = 2

√3i .


Limits and Continuity Continuity

Subsection 2

Continuity



Continuity of Real Functions

If the limit of a real function f as x approaches the point x0 existsand agrees with the value of the function f at x0, then we say that fis continuous at the point x0.

Continuity of a Real Function f (x)

A function f is continuous at a point x0 if limx→x0 f (x) = f (x0).

In order for the equation limx→x0 f (x) = f (x0) to hold:The limit limx→x0 f (x) must exist;f must be defined at x0;the two values must be equal.

If anyone of these three conditions fail, then f is not continuous at x0.

Example: The function f (x) =

{

x2, if x < 0x − 1, if x ≥ 0

is not continuous

at the point x = 0 since limx→0 f (x) does not exist.

Example: Even though limx→1x2−1x−1 = 2, the function f (x) = x2−1

x−1 isnot continuous at x = 1 because f (1) is not defined.



Continuity of Complex Functions

A complex function f is continuous at a point z0 if the limit of f as zapproaches z0 exists and is the same as the value of f at z0.

Definition (Continuity of a Complex Function)

A complex function f is continuous at a point z0 if limz→z0 f (z) = f (z0).

Criteria for Continuity at a Point

A complex function f is continuous at a point z0 if each of the followingthree conditions hold:

(i) limz→z0 f (z) exists;

(ii) f is defined at z0;

(iii) limz→z0 f (z) = f (z0).

If a complex function f is not continuous at a point z0, then we saythat f is discontinuous at z0.Example: The function f (z) = 1

1+z2is discontinuous at z = i and

z = −i .George Voutsadakis (LSSU) Complex Analysis October 2014 25 / 45


Checking Continuity at a Point

Consider the function f (z) = z2 − iz + 2.

To determine if f is continuous at the point z0 = 1− i , we must find

limz→z0 f (z);f (z0),and then check to see whether these two complex values are equal.

We obtain:limz→z0 f (z) = limz→1−i (z

2 − iz + 2) = (1−i)2−i(1−i)+2 = 1−3i .Furthermore, for z0 = 1− i , we have:f (z0) = f (1− i) = (1− i)2 − i(1− i) + 2 = 1− 3i .Since limz→z0 f (z) = f (z0), we conclude that f (z) = z2 − iz + 2 iscontinuous at the point z0 = 1− i .



Discontinuity of Principal Square Root Function

Show that the principal square root function f (z) = z1/2 =√

|z |e iArg(z)/2 is discontinuous at the point z0 = −1.

We show that the limit limz→z0 f (z) = limz→−1 z1/2 does not exist.

We let z approach −1 via two different paths.Consider z approaching −1 along the quarter of the unit circle lying inthe second quadrant, i.e., |z | = 1, π

2 < arg(z) < π. In exponential formz = e iθ, π

2 < θ < π, with θ approaching π. By setting |z | = 1 and

letting Arg(z) = θ approach π, we obtain: limz→−1 z1/2 =

limz→−1

√

|z |e iArg(z)/2 = limθ→π

√1e iθ/2 =

limθ→π (cosθ2 + i sin θ

2 ) = cos π2 + i sin π

2 = 0 + i(1) = i .Let z approach −1 along the quarter of the unit circle lying in the thirdquadrant, i.e., z = e iθ, −π < θ < −π

2 , with θ approaching −π. Bysetting |z | = 1 and letting Arg(z) = θ approach −π we find:limz→−1 z

1/2 = limz→−1

√

|z |e iArg(z)/2 = limθ→−π eiθ/2 =

limθ→−π (cosθ2 + i sin θ

2 ) = − i .

We conclude that limz→−1 z1/2 does not exist. Therefore,

f (z) = z1/2 is discontinuous at the point z0 = −1.



Continuity on a Set of Points

Besides continuity of a complex function f at a single point z0 in thecomplex plane, we are often also interested in the continuity of afunction on a set of points in the complex plane.

A complex function f is continuous on a set S if f is continuous atz0, for each z0 in S .

Example: Using the properties, we can show that f (z) = z2 − iz + 2is continuous at any point z0 in the complex plane. Therefore, we saythat f is continuous on C.

Example: The function f (z) = 1z2+1

is continuous on the setconsisting of all complex z such that z 6= ±i .



Real and Imaginary Parts of a Continuous Function

Various properties of complex limits can be translated into statementsabout continuity.

E.g., a preceding theorem described the connection between thecomplex limit of f (z) = u(x , y) + iv(x , y) and the real limits of u, v :

Definition (Continuity of a Real Function F (x , y))

A function F is continuous at (x0, y0) iflim(x ,y)→(x0,y0) F (x , y) = F (x0, y0).

Theorem (Real and Imaginary Parts of a Continuous Function)

Suppose that f (z) = u(x , y) + iv(x , y) and z0 = x0 + iy0. Then thecomplex function f is continuous at the point z0 if and only if both realfunctions u and v are continuous at the point (x0, y0).



Proof of the Theorem

Assume that the complex function f (z) = u(x , y) + iv(x , y) iscontinuous at z0 = x0 + iy0. Thenlimz→z0 f (z) = f (z0) = u(x0, y0) + iv(x0, y0). This implies:lim(x ,y)→(x0,y0) u(x , y) = u(x0, y0), lim(x ,y)→(x0,y0) v(x , y) = v(x0, y0).Therefore, both u and v are continuous at (x0, y0).

Conversely, if u and v are continuous at (x0, y0), thenlim(x ,y)→(x0,y0) u(x , y) = u(x0, y0) andlim(x ,y)→(x0,y0) v(x , y) = v(x0, y0). It then follows thatlimz→z0 f (z) = u(x0, y0) + iv(x0, y0) = f (z0). Therefore, f iscontinuous.



Checking Continuity Using the Theorem

Show that the function f (z) = z is continuous on C.

According to the theorem, f (z) = z = x + iy = x − iy is continuousat z0 = x0 + iy0 if both u(x , y) = x and v(x , y) = −y are continuousat (x0, y0).Because u and v are two-variable polynomial functions, it followsthat: lim(x ,y)→(x0,y0) u(x , y) = x0 and lim(x ,y)→(x0,y0) v(x , y) = −y0.This implies that u and v are continuous at (x0, y0). Therefore, f iscontinuous at z0 = x0 + iy0 by the preceding theorem.Since z0 = x0 + iy0 was arbitrary, we conclude that the functionf (z) = z is continuous on C.



Properties of Continuous Functions

Theorem (Properties of Continuous Functions)

If f and g are continuous at the point z0, then the following functions arecontinuous at the point z0:

(i) cf , c a complex constant;

(ii) f ± g ;

(iii) f · g ;(iv) f

g, provided g(z0) 6= 0.

We only prove (ii). Since f and g are continuous at z0, we have thatlimz→z0 f (z) = f (z0) and limz→z0 g(z) = g(z0). It follows thatlimz→z0 (f (z) + g(z)) = limz→z0 f (z)+ limz→z0 g(z) = f (z0)+ g(z0).Therefore, f + g is continuous at z0.



Continuity of Polynomial Functions

Theorem (Continuity of Polynomial Functions)

Polynomial functions are continuous on the entire complex plane C.

Let p(z) = anzn+ an−1z

n−1+ · · ·+ a1z + a0 be a polynomial functionand let z0 be any point in the complex plane C. The identity functionf (z) = z is continuous at z0, whence, by repeated application of theproduct rule, the power function f (z) = zn, where n is an integer andn ≥ 1, is continuous at this point as well. Moreover, every complexconstant function f (z) = c is continuous at z0, so it follows by thetheorem that each of the functions anz

n, an−1zn−1, . . ., a1z , and a0

are continuous at z0. Finally, from repeated application of the sumrule, p(z) = anz

n + an−1zn−1 + · · ·+ a1z + a0 is continuous at z0.

Since z0 was allowed to be any point in the complex plane, we haveshown that the polynomial function p is continuous on the entirecomplex plane C.



Continuity of Rational Functions

Continuity of Rational Functions

Rational functions are continuous on their domains.

Since a rational function f (z) = p(z)q(z) is quotient of the polynomial

functions p and q, it follows from the theorem and the quotient rulethat f is continuous at every point z0 for which q(z0) 6= 0.



Real and Complex Bounded Functions

Recall that if a real function f is continuous on a closed interval I onthe real line, then f is bounded on I , i.e., there is a real numberM > 0 such that |f (x)| ≤ M, for all x in I .

An analogous result for real functions F (x , y) states that, if F (x , y) iscontinuous on a closed and bounded region R of the Cartesian plane,then there is a real number M > 0, such that |F (x , y)| ≤ M, for all(x , y) in R , and we say F is bounded on R .

Suppose that the function f (z) = u(x , y) + iv(x , y) is defined on aclosed and bounded region R in the complex plane. As with realfunctions, we say that the complex function f is bounded on R ifthere exists a real constant M > 0, such that |f (z)| ≤ M, for all z inR .



Bounded Property for Complex Functions

Theorem (A Bounding Property)

If a complex function f is continuous on a closed and bounded region R ,then f is bounded on R . That is, there is a real constant M > 0, suchthat |f (z)| ≤ M, for all z in R .

If f is continuous on R , then u and v are continuous real functions onR . Since the square root function is continuous, it follows that thereal function F (x , y) =

√

u(x , y)2 + v(x , y)2 is also continuous on R .Because F is continuous on the closed and bounded region R , F isbounded on R , i.e., there is a real constant M > 0, such that|F (x , y)| ≤ M, for all (x , y) in R . However, since |f (z)| = F (x , y),we have that |f (z)| ≤ M, for all z in R . Thus, the complex functionf is bounded on R .



Branches

We have discussed the concept of a multiple valued function F (z)that assigns a set of complex numbers to the input z .

Examples of multiple valued functions include F (z) = z1/n, whichassigns to the input z the set of n n-th roots of z , and G (z) = arg(z),which assigns to the input z the infinite set of arguments of z .

In practice, it is often the case that we need a consistent way ofchoosing just one of the values of a multiple-valued function.

If we make this choice of value with the concept of continuity inmind, then we obtain a function that is called a branch of amultiple-valued function.

A branch of a multiple-valued function F is a function f1 that iscontinuous on some domain and that assigns exactly one of themultiple values of F to each point z in that domain.

Notation for Branches: When representing branches of a multiplevalued function F with functional notation, we will use lowercaseletters with a numerical subscript such as f1, f2, and so on.



Discontinuities of the Square Root Function

The requirement that a branch be continuous means that the domainof a branch is different from the domain of the multiple valuedfunction.

Example: The multiple-valued function F (z) = z1/2 that assigns toeach input z the set of two square roots of z is defined for all nonzerocomplex numbers z . Even though the principal square root functionf (z) = z1/2 does assign exactly one value of F to each input z (theprincipal square root of z), f is not a branch of F . The reason is thatthe principal square root function is not continuous on its domain.E.g., we showed that f (z) = z1/2 is not continuous at z0 = −1. Wecan also show that f (z) = z1/2 is discontinuous at every point on thenegative real axis.

In order to obtain a branch of F (z) = z1/2 that agrees with theprincipal square root function, we must restrict the domain to excludepoints on the negative real axis.



The Principal Branch of the Square Root Function

We define the principal branch of F (z) = z1/2 by f1(z) =√re iθ/2,

−π < θ < π. The function f1 is a branch of F (z) = z1/2.The domain Dom(f1) of f1 is defined by |z | > 0, −π < arg(z) < π.The function f1 agrees with the principal square root function f onthis set. Thus, f1 does assign to the input z exactly one of the valuesof F (z) = z1/2. To show that f1 is a continuous, let z be a point with|z | > 0, −π < arg(z) < π. If z = x + iy and x > 0, then z = re iθ,where r =

√

x2 + y2 and θ = tan−1 (yx). Since −π

2 < tan−1 (yx) < π

2 ,the inequality −π < θ < π is satisfied. Thus, substituting theexpressions for r and θ: f1(z) =

4√

x2 + y2e i tan−1 (y/x)/2 =

4√

x2 + y2 cos ( tan−1 (y/x)2 ) + i 4

√

x2 + y2 sin ( tan−1 (y/x)2 ). Because the

real and imaginary parts of f1 are continuous real functions for x > 0,we conclude that f1 is continuous for x > 0. A similar argument canbe made for points with y > 0 using θ = cot−1 ( x

y) and for points

with y < 0 using θ = − cot−1 ( xy). So f1 is continuous.



Branch Cuts

Although F (z) = z1/2 is defined for all nonzero complex numbers C,the principal branch f1 is defined only on |z | > 0, −π < arg(z) < π.

In general, a branch cut for a branch f1 of a multiple-valued functionF is a portion of a curve that is excluded from the domain of F sothat f1 is continuous on the remaining points.

Therefore, the non-positive real axis is a branch cut for the principalbranch f1 of the multiple-valued function F (z) = z1/2.

A different branch of F with the same branch cut is given byf2(z) =

√re iθ/2, π < θ < 3π. These are distinct since for, e.g., z = i ,

f1(i) =12

√2 + 1

2

√2i , but f2(i) = −1

2

√2− 1

2

√2i .

If we set φ = θ − 2π, then the branch f2 can be expressed asf2(z) =

√re i(φ+2π)/2 =

√re iφ/2e iπ, −π < φ < π. Since e iπ = −1,

f2(z) = −√re iφ/2, −π < φ < π. This shows that f2 = −f1.

These two branches of F (z) = z1/2 are analogous to the positive andnegative square roots of a positive real number.



Branch Points

The point z = 0 must be on the branch cut of every branch of themultiple-valued function F (z) = z1/2.

A point with the property that it is on the branch cut of every branchis called a branch point of F .

Alternatively, a branch point is a point z0 with the following property:

If we traverse any circle centered at z0 with sufficiently small radiusstarting at a point z1, then the values of any branch do not return tothe value at z1.

Example: Consider any branch of G (z) = arg(z).

At the point, say, z0 = 1, if we traverse thesmall circle |z − 1| = ε counterclockwise fromthe point z1 = 1 − εi , then the values of thebranch increase until we reach the point 1+εi .Then the values of the branch decrease backdown to the value of the branch at z1. Thus,z0 = 1 is not a branch point.



Example (Cont’d)

Consider again any branch of G (z) = arg(z).

Suppose the process is repeated for thepoint z0 = 0. For the small circle |z | = ε,the values of the branch increase along theentire circle. By the time we have re-turned to our starting point, the value ofthe branch is no longer the same, but hasincreased by 2π.

Therefore, z0 = 0 is a branch point of G (z) = arg(z).



Infinite Limits and Limits at Infinity

In analogy with real analysis, we can also define the concepts ofinfinite limits and limits at infinity for complex functions.

Intuitively, the limit limz→∞ f (z) = L means that values f (z) of thefunction f can be made arbitrarily close to L if values of z are chosenso that |z | is sufficiently large.

The limit of f as z tends to ∞ exists and is equal to L if, for everyε > 0, there exists a δ > 0, such that |f (z)− L| < ε whenever|z | > 1/δ.

Using this definition it is not hard to show that: limz→∞ f (z) = L ifand only if limz→0 f (

1z) = L.

Similarly, the infinite limit limz→z0 f (z) = ∞ is defined by:

The limit of f as z tends to z0 is ∞ if, for every ε > 0, there is aδ > 0, such that |f (z)| > 1/ε whenever 0 < |z − z0| < δ.

From this definition we obtain: limz→z0 f (z) = ∞ if and only iflimz→z0

1f (z) = 0.



Continuous Complex Parametric Curves

In real analysis we visualize a continuous function as a function whosegraph has no breaks or holes in it.

There is an analogous property for continuous complex functions, butit must be stated in terms of complex mappings.

A parametric curve defined by parametric equations x = x(t) andy = y(t) is called continuous if the real functions x , y are continuous.

Similarly, a complex parametric curve defined by z(t) = x(t) + iy(t)is continuous if both x(t) and y(t) are continuous real functions.

As with parametric curves in the Cartesian plane, a continuousparametric curve in the complex plane has no breaks or holes in it.

Such curves provide a means to visualize continuous complexfunctions.



Visualizing Continuity via Complex Parametric Curves

Proposition

If a complex function f is continuous on a set S , then the image of everycontinuous parametric curve in S must be a continuous curve.

Consider a continuous complex function f (z) = u(x , y) + iv(x , y) anda continuous parametric curve defined by z(t) = x(t) + iy(t). Wesaw that u(x , y) and v(x , y) are continuous real functions. Moreover,since x(t) and y(t) are continuous functions, it follows that thecompositions u(x(t), y(t)) and v(x(t), y(t)) are continuousfunctions. Therefore, the image of the parametric curve given by

w(t) = f (z(t)) = u(x(t), y(t)) + iv(x(t), y(t))

is also continuous.


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